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- Thread starter jason12345
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What is the definition of reducible? (I'm not sure what you mean.)

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What is the definition of reducible? (I'm not sure what you mean.)

It means it can be broken down into parts that transform among themselves.

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Can you provide the reference where the statement in your first post appears?

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Can you provide the reference where the statement in your first post appears?

It comes from Anderson's Principles of Relativty Physics on page 19:

" Thus from the components of a tensor T[tex]^{uv}[/tex] we can construct its symmetric part T[tex]^{(uv)}[/tex] and its antisymmetric part T[tex]^{[uv]}[/tex] according to

[tex]T^{(uv)} = 1/2(T^{uv}+ T^{vu})[/tex]

[tex]T^{[uv]} = 1/2(T^{uv}- T^{vu})[/tex]

Similarly we can construct the transformed symmetric part T[tex]^{'(uv)}[/tex] and the antisymmetric part T[tex]^{'[uv]}[/tex] from the transformed T[tex]^{'uv}[/tex]. Then, one can show that T[tex]^{'(uv)}[/tex] is a function of T[tex]^{(uv)}[/tex] and the mapping function only, and similarly for T[tex]^{[uv]}[/tex]. When ever a geometrical object can be broken up into parts that transform among themselves, we say that we have a reducible object. If no such decomposition as possible. we have an irreducible object.

- #6

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I take it you are referring to Problem 1.7....

where they are telling you to "decompose" [itex]T_\nu{}^\mu[/itex]

into irreducible parts consisting of

its "trace" [itex]T_\mu{}^\mu[/itex] and

a "traceless " (or "trace-free") tensor (which they give as [itex]T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho[/itex] on the assumption one is working in a 4-dimensional space).

So, basically they have told you to write [itex]T_\nu{}^\mu[/itex]

as a sum of two tensors, one of which they gave you:

[itex]T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu [/itex]

- #7

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I take it you are referring to Problem 1.7....

where they are telling you to "decompose" [itex]T_\nu{}^\mu[/itex]

into irreducible parts consisting of

its "trace" [itex]T_\mu{}^\mu[/itex] and

a "traceless " (or "trace-free") tensor (which they give as [itex]T_\nu{}^\mu-\frac{1}{4}\delta_\nu{}^\mu T_\rho{}^\rho[/itex] on the assumption one is working in a 4-dimensional space).

So, basically they have told you to write [itex]T_\nu{}^\mu[/itex]

as a sum of two tensors, one of which they gave you:

[itex]T_\nu{}^\mu = \mbox{(trace-part tensor)}_\nu{}^\mu + \mbox{(trace-free tensor)}_\nu{}^\mu [/itex]

Yes. The question says: "show that it is reducible *and* its irreducuble parts are...". So I would have thought you should be able to first show its reducible from the transformation properties of a mixed second rank tensor, rather than being told what the irreducible components are and from that showing its reducible :smile

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